# I'm a beginner in Calculus. I have 2 doubts.

### I'm a beginner in Calculus. I have 2 doubts.

Why cannot we take limit as Δy approaching 0 in the derivative formula?
Why do we not differentiate the coefficients with x? f.eg: y=3x^2=6x....shouldn't the constant=0??
Sorry if these are dumb.
Guest

### Re: I'm a beginner in Calculus. I have 2 doubts.

Guest wrote:Why cannot we take limit as Δy approaching 0 in the derivative formula?

??? Δy does go to 0! We don't say "as Δy goes to 0" because y is a function of x and what happens to y, so that Δy goes to zero, is because Δx goes to 0.

Why do we not differentiate the coefficients with x? f.eg: y=3x^2=6x....shouldn't the constant=0??
Sorry if these are dumb.

And we DO "differentuate the cofficients". But, as you say, the derivative of the constant is 0.
(Please do NOT write "Y= 3x^2= 6x". That says that 3x^2=6x which is not true! Similarly, the derivative of a constant is 0. "constant= 0" is not generally true!

Are you thinking that the derivative of 6x^2 should be (6)'(x^2)'= 0(2x)= 0? Since every expression has a ceffcient (even "x^2" has coefficient 1) the derivative of every thing would be 0! The derivative wouldn't be very useful!

If that is what you are thinking, you are missing the "product rule". The derivative of fg is f'g+ fg'. The derivative of 6x^2 is (6)'x^2+ 6(x^2)'= (0)x^2+ 6(2x)= 12x
Guest